We have $\Delta u=f$ in $D$, and $\dfrac{\partial u}{\partial n}+au=h$ on boundary of D, where $D$ is a domain in three dimension and $a$ is a positive constant. $\dfrac{\partial u}{\partial n}=\triangledown u\cdot n$ ($n$ is normal vector).
My thoughts: Suppose there are $u_1$ and $u_2$, satisfis the above equations. Let $w=u_1-u_2$, then we have $\Delta w=0$ in $D$, and $\dfrac{\partial w}{\partial n}=-aw$ on boundary of D. Maximum modulus principle may be useful but I don't know where to put it in. And energy method seems not helpful in this question.
Any help would be appreciated!
Best Answer
$\newcommand{div}{\nabla \cdot} \newcommand{grad}{\nabla}$ Start by taking $w$ as in your question so that $\Delta w = 0$ in $D$ and $\frac{\partial w}{\partial n} = -aw$ on $\partial D$ as you note.
Also note the following identity for the divergence operator for a scalar field $\phi$ and a vector field $\bf F$ $$\div (\phi {\bf F}) = \phi \div {\bf F} + \grad \phi \cdot {\bf F}$$
Then consider $$\iiint_D \div(w \grad w) dV = \iiint_D w \Delta w + (\grad w)^2 dV = \iiint_D (\grad w)^2 dV \geq 0$$ since $\Delta w = 0$ in $D$.
But also by the divergence theorem $$\iiint_D \div(w \grad w) dV = \iint_{\partial D} w \grad w \cdot {\bf dS} = \iint_{\partial D} w\grad w \cdot {\bf n} ds = \iint_{\partial D} -aw^2 ds \leq 0$$
Hence $$\iiint_D (\grad w)^2 dV = 0 \implies \grad w = 0 \mbox{ in } D$$ This gives us that $w$ is a constant so that solutions are unique up to a constant.